Derivada de Tan 2x

¿ Cuál es la Derivada de Tan 2x ?

La derivada de tan 2x es 2.(1+tan² 2x).

$(tan2x{)}^{\prime }=2.(1+ta{n}^{2}2x)$

$\frac{d}{dx}(tan2x)=2.(1+ta{n}^{2}2x)$

Prueba de la Derivada de Tan 2x

Método 1

$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$

$(tan2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{tan2(x+h)-tan2x}{h}$

$(tan2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{tan(2x+2h)-tan2x}{h}$

$\boxed{tan(p+q)=\frac{tanp+tanq}{1-tanp.tanq}}$

$(tan2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\frac{tan2x+tan2h}{1-tan2x.tan2h}-tan2x}{h}$

$(tan2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\frac{tan2x+tan2h-tan2x.(1-tan2x.tan2h)}{1-tan2x.tan2h}}{h}$

$(tan2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\frac{\cancel{tan2x}+tan2h-\cancel{tan2x}+ta{n}^{2}2x.tan2h}{1-tan2x.tan2h}}{h}$

$(tan2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\frac{tan2h+ta{n}^{2}2x.tan2h}{1-tan2x.tan2h}}{h}$

$(tan2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\frac{tan2h.(1+ta{n}^{2}2x)}{1-tan2x.tan2h}}{h}$

$(tan2x{)}^{\prime }=\underset{h\to 0}{\lim }[\frac{1}{h}.\frac{tan2h.(1+ta{n}^{2}2x)}{1-tan2x.tan2h}]$

$(tan2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{tan2h.(1+ta{n}^{2}2x)}{h.(1-tan2x.tan2h)}$

$(tan2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{2.tan2h.(1+ta{n}^{2}2x)}{2.h.(1-tan2x.tan2h)}$

$(tan2x{)}^{\prime }=2.\underset{h\to 0}{\lim }\frac{tan2h.(1+ta{n}^{2}2x)}{2h.(1-tan2x.tan2h)}$

$(tan2x{)}^{\prime }=2.\underset{h\to 0}{\lim }(\frac{tan2h}{2h}.\frac{1+ta{n}^{2}2x}{1-tan2x.tan2h})$

$(tan2x{)}^{\prime }=2.\underset{h\to 0}{\lim }\frac{tan2h}{2h}.\underset{h\to 0}{\lim }\frac{1+ta{n}^{2}2x}{1-tan2x.tan2h}$

$h\to 0(2h=h)$

$(tan2x{)}^{\prime }=2.\underset{h\to 0}{\lim }\frac{tanh}{h}.\underset{h\to 0}{\lim }\frac{1+ta{n}^{2}2x}{1-tan2x.tanh}$

$\boxed{\underset{t\to 0}{\lim }\frac{tant}{t}=1}$

$(tan2x{)}^{\prime }=\mathrm{2.1.}\frac{1+ta{n}^{2}2x}{1-tan2x.tan0}$

$\boxed{tan0=0}$

$(tan2x{)}^{\prime }=\mathrm{2.1.}\frac{1+ta{n}^{2}2x}{1-tan2x.0}$

$(tan2x{)}^{\prime }=\mathrm{2.1.}\frac{1+ta{n}^{2}2x}{1-0}$

$(tan2x{)}^{\prime }=\mathrm{2.1.}\frac{1+ta{n}^{2}2x}{1}$

$(tan2x{)}^{\prime }=\mathrm{2.1.}(1+ta{n}^{2}2x)$

$(tan2x{)}^{\prime }=2.(1+ta{n}^{2}2x)$

Método 2

$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$

$(tan2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{tan2(x+h)-tan2x}{h}$

$(tan2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{tan(2x+2h)-tan2x}{h}$

$\boxed{tanx=\frac{sinx}{cosx}}$

$(tan2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\frac{sin(2x+2h)}{cos(2x+2h)}-\frac{sin2x}{cos2x}}{h}$

$(tan2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\frac{sin(2x+2h).cos2x-cos(2x+2h).sin2x}{cos2x.cos(2x+2h)}}{h}$

$\boxed{sinp.cosq-cosp.sinq=sin(p-q)}$

$(tan2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\frac{sin(\cancel{2x}+2h-\cancel{2x})}{cos2x.cos(2x+2h)}}{h}$

$(tan2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\frac{sin2h}{cos2x.cos(2x+2h)}}{h}$

$(tan2x{)}^{\prime }=\underset{h\to 0}{\lim }[\frac{1}{h}.\frac{sin2h}{cos2x.cos(2x+2h)}]$

$(tan2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{sin2h}{h.cos2x.cos(2x+2h)}$

$(tan2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{2.sin2h}{2.h.cos2x.cos(2x+2h)}$

$(tan2x{)}^{\prime }=2.\underset{h\to 0}{\lim }\frac{sin2h}{2h.cos2x.cos(2x+2h)}$

$(tan2x{)}^{\prime }=2.\underset{h\to 0}{\lim }[\frac{sin2h}{2h}.\frac{1}{cos2x.cos(2x+2h)}]$

$(tan2x{)}^{\prime }=2.\underset{h\to 0}{\lim }\frac{sin2h}{2h}.\underset{h\to 0}{\lim }\frac{1}{cos2x.cos(2x+2h)}$

$h\to 0(2h=h)$

$(tan2x{)}^{\prime }=2.\underset{h\to 0}{\lim }\frac{sinh}{h}.\underset{h\to 0}{\lim }\frac{1}{cos2x.cos(2x+h)}$

$\boxed{\underset{t\to 0}{\lim }\frac{sint}{t}=1}$

$(tan2x{)}^{\prime }=\mathrm{2.1.}\frac{1}{cos2x.cos(2x+0)}$

$(tan2x{)}^{\prime }=\mathrm{2.1.}\frac{1}{cos2x.cos2x}$

$(tan2x{)}^{\prime }=\frac{\mathrm{2.1.1}}{co{s}^{2}2x}$

$(tan2x)=\frac{2}{co{s}^{2}2x}$

$(tan2x{)}^{\prime }=2.\frac{1}{co{s}^{2}2x}$

$(tan2x{)}^{\prime }=2.(\frac{1}{cos2x}{)}^2$

$\boxed{\frac{1}{cosx}=secx}$

$(tan2x{)}^{\prime }=2.se{c}^{2}2x$

$(tan2x{)}^{\prime }=2.(1+ta{n}^{2}2x)=\frac{2}{co{s}^{2}2x}=2.se{c}^{2}2x$